Show that a real \(2\times2\) matrix \(A\) satisfies \(A^2=-I\) if and only if \[ A=\begin{pmatrix} \pm\sqrt{pq-1}&-p\\ q&\mp\sqrt{pq-1} \end{pmatrix}, \] where \(p,q\in\mathbb R\), \(pq\ge1\), and the two signs are chosen consistently.
Show that for every \(\varepsilon>0\) there is no real \(2\times2\) matrix \(A\) such that \[ A^2=\begin{pmatrix}-1&0\\0&-1-\varepsilon\end{pmatrix}. \]
Solution
Write \[ A= \begin{pmatrix} a&b\\ c&d \end{pmatrix}. \]
1. If \(A^2=-I\), then\[ a+d=0. \]
Expanding the square gives \[ A^2= \begin{pmatrix} a^2+bc&b(a+d)\\ c(a+d)&d^2+bc \end{pmatrix}. \] Thus \(A^2=-I\) implies \[ a^2+bc=-1, \qquad d^2+bc=-1, \] and hence \[ a^2=d^2. \] If \(a+d\neq0\), the off-diagonal equations \[ b(a+d)=0, \qquad c(a+d)=0 \] would force \(b=c=0\). The first diagonal equation would then give \(a^2=-1\), impossible over \(\RR\). Therefore \(a+d=0\).
2. If \(A^2=-I\), then there are \(p,q\in\RR\) with \(pq\ge1\) suchthat\[ A= \begin{pmatrix} \pm\sqrt{pq-1}&-p\\ q&\mp\sqrt{pq-1} \end{pmatrix}. \]
By step 1, \(d=-a\). The diagonal equation becomes \[ a^2+bc=-1. \] Set \[ p=-b, \qquad q=c. \] Then \[ a^2-pq=-1, \] so \[ pq=a^2+1\ge1 \] and \[ a=\pm\sqrt{pq-1}. \] Since \(d=-a\), the two diagonal signs are opposite, exactly as in the stated form.
3. Every matrix of the displayed form in part 1 satisfies\[ A^2=-I. \]
Let \[ s=\pm\sqrt{pq-1}, \qquad A= \begin{pmatrix} s&-p\\ q&-s \end{pmatrix}. \] Then \[ A^2= \begin{pmatrix} s^2-pq&0\\ 0&s^2-pq \end{pmatrix}. \] Since \(s^2=pq-1\), this is \(-I\).
4. Part 1 is therefore proved.
Step 2 proves necessity, and step 3 proves sufficiency.
5. For part 2, suppose for contradiction that a real matrix \(A\)satisfies\[ A^2= D \coloneqq \begin{pmatrix} -1&0\\ 0&-1-\varepsilon \end{pmatrix}, \qquad \varepsilon>0. \]Then \(A\) commutes with \(D\).
Since \(D=A^2\), \[ AD = AA^2 = A^3 = A^2A = DA. \]
6. Any real \(2\times2\) matrix commuting with \(D\) is diagonal.
Write \[ A= \begin{pmatrix} a&b\\ c&d \end{pmatrix}. \] The equation \(AD=DA\) gives \[ -b(1+\varepsilon)=-b, \qquad -c=-c(1+\varepsilon). \] Since \(\varepsilon>0\), these equations force \[ b=c=0. \] Thus \(A\) is diagonal.
7. No real diagonal matrix can square to \(D\).
If \[ A= \begin{pmatrix} a&0\\ 0&d \end{pmatrix}, \] then \[ A^2= \begin{pmatrix} a^2&0\\ 0&d^2 \end{pmatrix}. \] Equality with \(D\) would require \(a^2=-1\), impossible for \(a\in\RR\).
8. Therefore, for every \(\varepsilon>0\), there is no real\(2\times2\) matrix whose square is\[ \begin{pmatrix} -1&0\\ 0&-1-\varepsilon \end{pmatrix}. \]
Steps <1>5–<1>7 show that assuming such a matrix exists leads to a contradiction.
9. Q.E.D.
Step 4 proves part 1, and step 8 proves part 2.